I recently came across this question:

Is the axiom of choice needed to prove the following statement:

Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. There exists a linear map $T : V \rightarrow W$ such that $Tv = w$.

I have talked to someone, who went and asked a few people, and they think that just ZF by itself is not sufficient, and that we indeed do need AC. Furthermore, they think that the statement is itself not sufficient to prove AC. Can anyone give a definitive answer?